# Two persons A and B are throwing an unbiased six faced die alternatively, with the condition that the person who throws $3$ first wins the game. If A starts the game, the probabilities of A and B to win the game are respectively

$\left( a \right)\dfrac{6}{{11}},\dfrac{5}{{11}}$

$\left( b \right)\dfrac{5}{{11}},\dfrac{6}{{11}}$

$\left( c \right)\dfrac{8}{{11}},\dfrac{3}{{11}}$

$\left( d \right)\dfrac{3}{{11}},\dfrac{8}{{11}}$

Last updated date: 22nd Mar 2023

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Answer

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Hint: Use formula of sum of infinite geometric series $S = \dfrac{a}{{1 - r}}$

Winning the game is getting a $3$ on the die $p\left( {getting 3} \right) = \dfrac{1}{6}$ and $q\left( {not getting 3} \right) = \dfrac{5}{6}$

A wins if he gets $3$ on his first turn or he gets $3$ on his second turn but B does not get $3$ on his first turn and so on.

$

p\left( A \right) = p + p{q^2} + p{q^4} + ....... \\

\Rightarrow p\left( {1 + {q^2} + {q^4} + ........} \right) \\

$

We can see sum of infinite geometric series formed of common ratio $r = {q^2}$and first term $a = 1$ then sum of infinite geometric series is $S = \dfrac{a}{{1 - r}}$

$\left( {1 + {q^2} + {q^4} + ........} \right) = \dfrac{1}{{1 - {q^2}}}$

$

\Rightarrow p\left( {1 + {q^2} + {q^4} + ........} \right) \\

\Rightarrow p\left( {\dfrac{1}{{1 - {q^2}}}} \right) \\

\Rightarrow \dfrac{p}{{1 - {q^2}}} \\

\Rightarrow \dfrac{{\dfrac{1}{6}}}{{1 - {{\left( {\dfrac{5}{6}} \right)}^2}}} \Rightarrow \dfrac{6}{{11}} \\

p\left( A \right) = \dfrac{6}{{11}} \\

p\left( B \right) = 1 - p\left( A \right) = 1 - \dfrac{6}{{11}} \\

p\left( B \right) = \dfrac{5}{{11}} \\

$

Probability of A to win the game is $p\left( A \right) = \dfrac{6}{{11}}$ .

Probability of B to win the game is $p\left( B \right) = \dfrac{5}{{11}}$.

So, the correct option is (A).

Note: Whenever we come across these types of problems first find the probability of winning or losing the game in the first attempt but we know it is not possible to win the game in the first attempt. So, we try unless anyone is not winning the game then we use a sum of infinite geometric series.

Winning the game is getting a $3$ on the die $p\left( {getting 3} \right) = \dfrac{1}{6}$ and $q\left( {not getting 3} \right) = \dfrac{5}{6}$

A wins if he gets $3$ on his first turn or he gets $3$ on his second turn but B does not get $3$ on his first turn and so on.

$

p\left( A \right) = p + p{q^2} + p{q^4} + ....... \\

\Rightarrow p\left( {1 + {q^2} + {q^4} + ........} \right) \\

$

We can see sum of infinite geometric series formed of common ratio $r = {q^2}$and first term $a = 1$ then sum of infinite geometric series is $S = \dfrac{a}{{1 - r}}$

$\left( {1 + {q^2} + {q^4} + ........} \right) = \dfrac{1}{{1 - {q^2}}}$

$

\Rightarrow p\left( {1 + {q^2} + {q^4} + ........} \right) \\

\Rightarrow p\left( {\dfrac{1}{{1 - {q^2}}}} \right) \\

\Rightarrow \dfrac{p}{{1 - {q^2}}} \\

\Rightarrow \dfrac{{\dfrac{1}{6}}}{{1 - {{\left( {\dfrac{5}{6}} \right)}^2}}} \Rightarrow \dfrac{6}{{11}} \\

p\left( A \right) = \dfrac{6}{{11}} \\

p\left( B \right) = 1 - p\left( A \right) = 1 - \dfrac{6}{{11}} \\

p\left( B \right) = \dfrac{5}{{11}} \\

$

Probability of A to win the game is $p\left( A \right) = \dfrac{6}{{11}}$ .

Probability of B to win the game is $p\left( B \right) = \dfrac{5}{{11}}$.

So, the correct option is (A).

Note: Whenever we come across these types of problems first find the probability of winning or losing the game in the first attempt but we know it is not possible to win the game in the first attempt. So, we try unless anyone is not winning the game then we use a sum of infinite geometric series.

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